Commutator of an abelian normal subgroup and the cyclic group generated by x I'm working through the book "The Theory of Finite Groups" by Hans Kurzweil and Bernd Stellmacher.  Section 1.6, Exercise 1 is: 

Let $A$ be an abelian normal subgroup of $G$ and $x \in G$.  Show $[A, \langle x \rangle]$ = $\{[a, x] \mid a \in A\}$.  

Defns:
Commutator: $ [x, y]=x^{-1} y^{-1} x y$
$[A, \langle x \rangle]= \langle \{[a, x^i] \mid a \in A, x^i \in \langle x \rangle\} \rangle$
As far as I've gotten:
Obviously {$[a, x] | a \in A$} is contained in $ [A, <x>]. $ 
I need to show that $[a, x^i] = [a', x] $ for some $a' \in A.$  I've tried $a'= [a, x^{i-1}], $ and a number of other variations, but I can't find the trick that makes it work.  Thank you for any help.  
 A: (The following is adapted from my answer to the closely related question Show that the commutator subgroup of an abelian normal subgroup and whole group has special form..)
We may assume that the group $G$ is generated by $A$ and $x$ (if not, replace $G$ with the subgroup generated by $A$ and $x$).  Let $$H=\{[a,x]:a\in A\}.$$  First, let's show that the normal subgroup generated by $H$ contains $[A,\langle x\rangle]$.  That is, we want to show that if we apply any homomorphism $\varphi$ on $G$ that kills $H$, $\varphi(a)$ commutes with $\varphi(y)$ for any $a\in A$, $y\in\langle x\rangle$.  Since $\varphi$ kills $H$, $\varphi(a)$ commutes with $\varphi(x)$ for any $a\in A$.  Since any $y\in\langle x\rangle$ is a power of $x$, this implies $\varphi(a)$ commutes with any such $\varphi(y)$, as desired.
It now suffices to show that $H$ is a normal subgroup.  First, to show that $H$ is a subgroup, observe that $[a,x]=a(xa^{-1}x^{-1})=f(a)g(a)^{-1}$ where $f(a)=a$ and $g(a)=xax^{-1}$.  Both $f$ and $g$ are homomorphisms $A\to A$ ($g(a)\in A$ since $A$ is normal).  Since $A$ is abelian, a difference of two homomorphisms to $A$ is again a homomorphism.  So, $H$ is just the image of $A$ under the homomorphism $h:A\to A$ defined by $h(a)=[a,x]$, and so is a subgroup.
For normality, fix $a\in A$ and $y\in G$.  Since $G$ is generated by $A$ and $x$ and $A$ is normal in $G$, we can write $y=bx^i$ for some $i\in\mathbb{Z}$.  Then $$y[a,x]y^{-1}=b[x^iax^{-i},x^ixx^{-i}]b^{-1}=b[x^iax^{-i},x]b^{-1}.$$
Since $A$ is normal $x^iax^{-i}\in A$, and hence $[x^iax^{-i},x]\in A$ as well, so it commutes with $b$ since $A$ is abelian.  Thus $y[a,x]y^{-1}=[x^iax^{-i},x]\in H$.  Since $y\in G$ and $a\in A$ were arbitrary, $H$ is normal.
